$ C = \left[\begin{array}{rr}-1 & 2 \\ 5 & 3\end{array}\right]$ $ E = \left[\begin{array}{rr}5 & 1 \\ -2 & -1\end{array}\right]$ What is $ C E$ ?
Solution: Because $ C$ has dimensions $(2\times2)$ and $ E$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C E = \left[\begin{array}{rr}{-1} & {2} \\ {5} & {3}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{1} \\ {-2} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{2}\cdot{-2} & ? \\ {5}\cdot{5}+{3}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{5}+{2}\cdot{-2} & {-1}\cdot\color{#DF0030}{1}+{2}\cdot\color{#DF0030}{-1} \\ {5}\cdot{5}+{3}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{5}+{2}\cdot{-2} & {-1}\cdot\color{#DF0030}{1}+{2}\cdot\color{#DF0030}{-1} \\ {5}\cdot{5}+{3}\cdot{-2} & {5}\cdot\color{#DF0030}{1}+{3}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-9 & -3 \\ 19 & 2\end{array}\right] $